# Suppose a laboratory has a 31 g sample of polonium-210. The half-life of polonium-210 is about 138 days. How many half lives of polonium occur in 966 days? How much polonium is in the sample 966 days later?

May 19, 2017

$7$ half-lives

$0.24 g$ "^(210)Po

#### Explanation:

The number of half-lives is simply the number of days given ($966 d$) divided by the half-life ($138 d$), which is $7$ half-lives.

The amount of "^(210)Po remaining after $t$ days is represented by the equation

N(t) = N_0(1/2)^(t/(t_(1/2))

where $N \left(t\right)$ is the amount of substance remaining after $t$ days,
${N}_{0}$ is the initial amount of substance,
$t$ is the time in days, and
${t}_{\frac{1}{2}}$ is the half-life of the substance.

Plugging in known variables, we have

$N \left(t\right) = \left(31 g\right) {\left(\frac{1}{2}\right)}^{\frac{966}{138}} = 0.24 g$ "_(210)Po